Abstract

Let $A$, $B$ be two regular commutative unital Banach algebras such that $B$ is integral over $A$. In 2003, Dawson and Feinstein showed that the topological stable rank $\operatorname{tsr}(B) = 1$ whenever $\operatorname{tsr}(A) = 1$. In this note, we investigate whether we will have $\operatorname{tsr}(A) = \operatorname{tsr}(B)$ in general. For instance, when $A$ is a commutative unital $C^*$-algebra, we show that $\operatorname{tsr}(A) \leq \operatorname{tsr}(B)$, and the equality holds at least when the integral extension is separable. In general, $A$ and $B$ have the same Bass stable ranks $\operatorname{Bsr}(A) = \operatorname{Bsr}(B)$.

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